As we know in 3-Stage RC oscillator phase shift the frequency of feedback network is given by f=1/2*pi*R*C*sqrt (2*N) where N is number of stages
in 3-stage the frequency f=1/2*pi*R*C*sqrt (2*3)=1/2*pi*R*C*sqrt (6)
so when i try to get the frequency of 4 stages using same steps (KCL ) I don't get the expected result of f = 1/2*pi*RC*sqrt(8) . anyone can help ??

 

Hi, show your work.

 

Hi, show your work.

Thanks in advance

 

View attachment 307943
As we know in 3-Stage RC oscillator phase shift the frequency of feedback network is given by f=1/2*pi*R*C*sqrt (2*N) where N is number of stages
in 3-stage the frequency f=1/2*pi*R*C*sqrt (2*3)=1/2*pi*R*C*sqrt (6)
so when i try to get the frequency of 4 stages using same steps (KCL ) I don't get the expected result of f = 1/2*pi*RC*sqrt(8) . anyone can help ??

You may have found the statement I've shown in red at a web site like this: https://www.electronics-tutorials.ws/oscillator/rc_oscillator.html

That is a mistake; it is only true for N=3. For N=4 or more you will have to work it out for your self, or find the result elsewhere on the web.

 

Thanks in advance

It looks like you have the correct transfer function, but you haven't got a result for the frequency of oscillation. What do you get for it?

 

As we know in 3-Stage RC oscillator phase shift the frequency of feedback network is given by f=1/2*pi*R*C*sqrt (2*N) where N is number of stages
in 3-stage the frequency f=1/2*pi*R*C*sqrt (2*3)=1/2*pi*R*C*sqrt (6)
so when i try to get the frequency of 4 stages using same steps (KCL ) I don't get the expected result of f = 1/2*pi*RC*sqrt(8) . anyone can help ??

No that is not correct.
The expression you have mentioned applies for a three-stage RC oscillator only (leading to SQRT(6) )
More than that, the shown phase values (45 deg) are wrong.

At first, such a network works as a whole only; you cannot divide the total phase shift equally between the individual stages.
In addition, the oscillator frequency results from the condition that the total phase shift must be 180 degrees.
For this reason, at least three RC elements are required to obtain this phase condition at a finite frequency.

 

View attachment 307943
As we know in 3-Stage RC oscillator phase shift the frequency of feedback network is given by f=1/2*pi*R*C*sqrt (2*N) where N is number of stages
in 3-stage the frequency f=1/2*pi*R*C*sqrt (2*3)=1/2*pi*R*C*sqrt (6)
so when i try to get the frequency of 4 stages using same steps (KCL ) I don't get the expected result of f = 1/2*pi*RC*sqrt(8) . anyone can help ??

In a 4-stage RC oscillator, the phase shift is not equally distributed over the 4 stages(sections) of the RC network as shown in the figure. This is due to the loading of the first section by the following 3 sections, the loading of the 2nd section by the following 2 sections, and the loading of the 3rd section by the last section. When the Overall phase shift is 180 degrees, the cumulative phase shift starting from left is not 45, 90, 135 and 180 degrees.
In a manual analysis that I performed for a 3-stage low pass (series R and shunt C) RC feedback network, when the overall phase shift is -180 degrees, the phase shift at the first node is -55.8 degrees, the phase shift at the second node is -112.2 degrees (and the final phase shift is -180 degrees).

 

Look at the BUBBA oscillator that uses an opamp between each of the RC sections:

 

Look at the BUBBA oscillator that uses an opamp between each of the RC sections:

This BUBBA oscillator is - perhaps - easier to calculate.
But I think this is the only "advantage" (for the price of three additional opamps).

 

The BUBBA oscillator uses a quad opamp that is inside one package.

 

The BUBBA oscillator uses a quad opamp that is inside one package.

Yes - thats true. However, is there any additional advantage if compared with the one-opamp-version?

 

Bubba Oscillator The bubba oscillator in Figure 18 is another phase-shift oscillator, but it takes advantage of the quad op-amp package to yield some unique advantages. Four RC sections require 45° phase shift per section, so this oscillator has an excellent dφ/dt resulting in minimal frequency drift. The RC sections each contribute 45° phase shift, so taking outputs from alternate sections yields low-impedance quadrature outputs. When an output is taken from each op amp, the circuit delivers four 45 phase-shifted sine waves.

From the PDF...

 

The bubba oscillator........some unique advantages..... this oscillator has an excellent dφ/dt resulting in minimal frequency drift.

Please, can you explain why this oscillator topology has better frequency drift properties than the one-opamp structure?

 

Without the BUBBA opamp buffers, each RC section "loads down" the previous RC section requiring different calculations and much more opamp gain causing more distortion.

 

Without the BUBBA opamp buffers, each RC section "loads down" the previous RC section requiring different calculations and much more opamp gain causing more distortion.

Comparing both alternatives (one-opamp version vs. BUBBA) we see that the required gain values are 8 resp. 4.
So - I do not think that there will be a remarkable difference in distortion.
The most important contribution in distortion will be, of course, caused by the amplitude control mechanism.
However, regarding accuracy of the calculated oscillation frequency, there is one big disadvantage of the BUBBA type - not yet mentioined:
The input resistance of the inverter must be chosen large enough not to load the last RC-section.
Of course - improvements are possible.

May I repeat my previous question?
Why do you think that the BUBBA topology has better frequency drift properties than the one-opamp structure?

 

Comparing both alternatives (one-opamp version vs. BUBBA) we see that the required gain values are 8 resp. 4.
So - I do not think that there will be a remarkable difference in distortion.
The most important contribution in distortion will be, of course, caused by the amplitude control mechanism.
However, regarding accuracy of the calculated oscillation frequency, there is one big disadvantage of the BUBBA type - not yet mentioined:
The input resistance of the inverter must be chosen large enough not to load the last RC-section.
Of course - improvements are possible.

May I repeat my previous question?
Why do you think that the BUBBA topology has better frequency drift properties than the one-opamp structure?

Hello there,

I myself would question the veracity of that Pdf paper there are some questionable things I see there.
For one, who would take the output from the oscillator from across one of the R's of one of the RC sections? Not me that's for sure Take the output from one of the op amp outputs.
Also, I don't like the fact that one of the RC sections is loaded by the inverting gain stage. That 'could' complicate the simplicity of using four op amp sections.
I did not go over the gain requirements yet but I have a feeling they did not get it right. The single op amp version will require more gain though in any case.

The advantage of lower gain however is of course frequency response and that could lead to less distortion because there would be more gain left over for the lower frequencies of oscillations.

Another advantage is the ability to get both sine and cosine outputs. However, since one of the phase shifts in the single op amp circuit is close to 90 degrees already, it is most likely possible to alter some of the resistor values in the RC sections in order to get this feature in the single op amp circuit also. Remember in all these designs so far they have only considered using resistors of all the same value and capacitors of all the same value. If we alter that, we could get different values for the phase shifts of each section. Probably not too hard to calculate just a little more work. This may or may not alter the gain, but if so, just a little I would imagine. A buffer would probably still be required though for that second output. This could actually lead to a two op amp hybrid oscillator.

One of the main advantages I have found over the years is the multiple op amp version is a good starter for teaching the theory on how phase shift oscillators work. That is of course because of the ability to do a quick calculation rather than muddy up the theory with some complicated stage interactions which requires a more careful analysis, which itself could end up being in error and thus complicating the entire learning experience.

The question of drift is an interesting one. This could be because in the multiple op amp version each stage contributes one increment of phase and amplitude drift per stage resulting in a sum of incremental drifts of all the stages (4x). With the single op amp version the last stage drift is affected by the second to last stage drift as well as itself. The second to last stage would be affected by itself as well as the previous two stages, etc. This seems like we would have the last stage drift plus the effect of three more stages, plus the second to last stage drift plus the effect of two more stages, plus the third to last stage drift plus the effect of one more stage, plus the fourth stage drift. In units of incremental drift, this could be (1+3)+(1+2)+(1+1)+(1)=10x. Thus, we may see increased temperature drift due to the interaction of the stages of more than two times.
This should be verified, however, as this assumes equal drift interactions which would not be the case, stages farther apart would have less effect on each other. The effects would still be larger than the single op amp version though.

 

View attachment 307943
As we know in 3-Stage RC oscillator phase shift the frequency of feedback network is given by f=1/2*pi*R*C*sqrt (2*N) where N is number of stages
in 3-stage the frequency f=1/2*pi*R*C*sqrt (2*3)=1/2*pi*R*C*sqrt (6)
so when i try to get the frequency of 4 stages using same steps (KCL ) I don't get the expected result of f = 1/2*pi*RC*sqrt(8) . anyone can help ??

Hi,

As others have said, the phase shifts do not add up like that. They have the same phase shifts but only in the multiple op amp version of a phase shift oscillator.
I see you got the transfer function right but now you have to calculate the operating frequency and the required gain. Do you know how to do this?

 

View attachment 307943
As we know in 3-Stage RC oscillator phase shift the frequency of feedback network is given by f=1/2*pi*R*C*sqrt (2*N) where N is number of stages
in 3-stage the frequency f=1/2*pi*R*C*sqrt (2*3)=1/2*pi*R*C*sqrt (6)
so when i try to get the frequency of 4 stages using same steps (KCL ) I don't get the expected result of f = 1/2*pi*RC*sqrt(8) . anyone can help ??

The formula for the frequency is correct and it has a sqrt(6) factor in the denominator for the 3-stage RC oscillator, but that cannot be generalized to 6=2*3 and then use the formula f = 1/2*pi*RC*sqrt(8) for a 4-stage RC phase shift oscillator. The 4-stage RC phase shift oscillator formula is much different from the 3-stage RC phase shift oscillator formula.

 

View attachment 307943
As we know in 3-Stage RC oscillator phase shift the frequency of feedback network is given by f=1/2*pi*R*C*sqrt (2*N) where N is number of stages
in 3-stage the frequency f=1/2*pi*R*C*sqrt (2*3)=1/2*pi*R*C*sqrt (6)
so when i try to get the frequency of 4 stages using same steps (KCL ) I don't get the expected result of f = 1/2*pi*RC*sqrt(8) . anyone can help ??

Hello again,

I am not sure where you got this formula f=1/(2*pi*RC*sqrt(8)) that is not right for a regular direct coupled four stage RC filter, and it's not right for a buffered version either.

For this circuit the right formula is:
w=sqrt(7)/(sqrt(10)*RC) which is about 0.83666/RC,
or:
f=sqrt(7)/(2*pi*sqrt(10)*RC)

so it is no wonder why you did not get the expression with the sqrt(8) in the denominator.


For a 5 stage circuit it would be:
w=sqrt(14-sqrt(181))/(sqrt(15)*RC)
which is about 0.19085/RC.

 

Hello again,

I am not sure where you got this formula f=1/(2*pi*RC*sqrt(8)) that is not right for a regular direct coupled four stage RC filter, and it's not right for a buffered version either.

I explained where he likely got that in post 4.